Cutting Boards
Problem Statement :
Alice gives Bob a board composed of wooden squares and asks him to find the minimum cost of breaking the board back down into its individual squares. To break the board down, Bob must make cuts along its horizontal and vertical lines. To reduce the board to squares, Bob makes horizontal and vertical cuts across the entire board. Each cut has a given cost, or for each cut along a row or column across one board, so the cost of a cut must be multiplied by the number of segments it crosses. The cost of cutting the whole board down into squares is the sum of the costs of each successive cut. Can you help Bob find the minimum cost? The number may be large, so print the value modulo . For example, you start with a board. There are two cuts to be made at a cost of for the horizontal and for the vertical. Your first cut is across piece, the whole board. You choose to make the horizontal cut between rows and for a cost of . The second cuts are vertical through the two smaller boards created in step between columns and . Their cost is . The total cost is and . Function Description Complete the boardCutting function in the editor below. It should return an integer. boardCutting has the following parameter(s): cost_x: an array of integers, the costs of vertical cuts cost_y: an array of integers, the costs of horizontal cut Input Format The first line contains an integer , the number of queries. The following sets of lines are as follows: The first line has two positive space-separated integers and , the number of rows and columns in the board. The second line contains space-separated integers cost_y[i], the cost of a horizontal cut between rows and of one board. The third line contains space-separated integers cost_x[j], the cost of a vertical cut between columns and of one board.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define LLU unsigned long long int
typedef struct {
LLU data;
int side;
} point;
int my_compare(const void *i1, const void *i2)
{
point *p1 = (point *)i1;
point *p2 = (point *)i2;
if(p1->data > p2->data) return -1;
else return 1;
}
int main() {
int t;
LLU answer=0, seg_count[2], m, n, i;
LLU mod = (LLU)pow(10, 9) + 7;
point *points, cur_point;
//printf("mod=%llu %d %d\n", mod, sizeof(int), sizeof(LLU));
scanf("%d",&t);
while(t--) {
scanf("%llu %llu", &m, &n);
answer = 0;
seg_count[0] =1;
seg_count[1] =1;
points = (point *)malloc(sizeof(point)*(m+n-2));
for(i=0;i<m-1;i++) {
scanf("%llu", &points[i].data);
points[i].side = 1;
}
for(i=0;i<n-1;i++) {
scanf("%llu", &points[i+m-1].data);
points[i+m-1].side = 0;
}
qsort(points, m+n-2, sizeof(point), my_compare);
for(i=0;i<m+n-2;i++) {
cur_point = points[i];
switch(cur_point.side) {
case 1:
answer+= (seg_count[1] % mod) * (cur_point.data % mod);
seg_count[0]++;
break;
case 0:
answer+= (seg_count[0] % mod) * (cur_point.data % mod);
seg_count[1]++;
break;
}
answer = answer % mod;
}
printf("%llu\n", answer);
free(points);
}
return 0;
}
Solution in C++ :
In C++ :
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
bool compare(pair<char,unsigned long long> A,pair<char,unsigned long long> B)
{
return A.second > B.second;
}
int main()
{
int T;
cin>>T;
for(;T--;)
{
int M,N,cost;
cin>>M>>N;
vector<pair<char,unsigned long long> > cutCost;
for (int i = 0; i < M-1; ++i)
{
cin>>cost;
cutCost.push_back(make_pair('y',cost));
}
for (int i = 0; i < N-1; ++i)
{
cin>>cost;
cutCost.push_back(make_pair('x',cost));
}
sort(cutCost.begin(),cutCost.end(),compare);
unsigned long long vcut = 1, hcut = 1, totalcost = 0;
for (int i = 0; i < cutCost.size(); ++i)
{
if(cutCost[i].first == 'y')
{
totalcost = (totalcost + ((vcut*cutCost[i].second)%1000000007))%1000000007;
++hcut;
}
else if(cutCost[i].first == 'x')
{
totalcost = (totalcost + ((hcut*cutCost[i].second)%1000000007))%1000000007;
++vcut;
}
}
cout<<totalcost<<"\n";
}
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(new BufferedInputStream(System.in));
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
int m = sc.nextInt();
int n = sc.nextInt();
Integer[] yi = new Integer[m-1];
Integer[] xi = new Integer[n-1];
for(int j=0;j<m-1;j++){
yi[j]= sc.nextInt();
}
for(int j=0;j<n-1;j++){
xi[j]= sc.nextInt();
}
Arrays.sort(yi,Collections.reverseOrder());
Arrays.sort(xi,Collections.reverseOrder());
int ny=1,nx=1;
long c=0;
while(ny<m || nx<n) {
if(ny<m && (nx>=n || yi[ny-1]>xi[nx-1])) {
c= (c + ((long)nx*(long)yi[ny-1])%1000000007)%1000000007;
ny++;
} else if(nx<n && (ny>=m || xi[nx-1]>=yi[ny-1])) {
c= (c + ((long)ny*(long)xi[nx-1])%1000000007)%1000000007;
nx++;
}
}
System.out.println(c);
}
}
}
Solution in Python :
In Python3 :
for t in range(int(input())):
m,n = map(int,input().split())
hor = sorted(map(int,input().split()),reverse=True)
ver = sorted(map(int,input().split()),reverse=True)
h = v = 0
ret = 0
modulo = 10**9 + 7
for i in range(m+n):
if h>=len(hor) or v>=len(ver): break
if ver[v] > hor[h] :
ret += ((h+1)*ver[v])%modulo
v += 1
else:
ret += (hor[h] *(v+1)) % modulo
h += 1
if h<len(hor):
ret += (sum(hor[h:])*(v+1)) % modulo
elif v<len(ver):
ret += (sum(ver[v:])*(h+1)) % modulo
print(ret% modulo)
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